Newest questions tagged geometry - Physics Stack Exchange

Rate of change of surface area is acting weird

2024-08-23T04:43:11Z

Preface- I made this situation up. An uniform metal sphere of given mass $M$, is being continuously fed more mass at its centre at the rate $4 \ \text{kg/s}$. If the density stays constant, find the rate of increase in surface area.

How I tried to solve this $$\rho V=M\implies \rho \frac{dV}{dt}=\frac{dM}{dt}$$ So $$4π\rho r^2 \frac{dr}{dt}=4\implies \frac{dr}{dt}=\frac{1}{π\rho r^2}$$ From here we can easily derive that $\frac{dA}{dt}=\frac{8}{\rho r}$ using $A=4πr^2$

But we see that rate of increase in surface area is inversely proportional to that of the radius, so if radius increases, will the surface area increase at a slower rate? I can't actually visualise this, and so I can't analyse or make a diagram

Are my methods wrong?

Is there a general solution to all spherical triangles as described by Arnold Sommerfeld?

2024-08-20T05:42:20Z

Arnold Sommerfeld has demonstrated that it is legal to use spherical trigonometry in solving relative velocity compositions. In this work,

https://en.wikisource.org/wiki/Translation:On_the_Composition_of_Velocities_in_the_Theory_of_Relativity

he shows his solution (restricted to two right angled congruent triangles) is equivalent to the Lorentz transform.

I am looking for a solution to the general case i.e. all spherical triangles whether right angled or congruent or not.

What is the line of sight distance across the ocean? [closed]

2024-08-20T02:18:49Z

I just watched an experiment where they had a laser at a height of 50 feet (15 metre) above sea level and were able to see it 23 miles (37 km) away at a receiver which was 20 feet (6 metre) above sea level. According to this chart:

http://www.totally-cuckoo.com/distance_visible_to_the_horizon.htm

It says the distance to the horizon at 50 feet (15 metre) is 9.35 miles (15 km) and the distance at 20 feet (6 metre) is 5.92 miles (9.5 km). That's 15.27 miles (24.6 km) total so how was the laser seen at 23 miles (37 km)? Even with a refraction rate of 8% that's only 1.84 miles (3.0 km) for a total of 17.11 miles (27.5 km). So how was the laser seen at 23 miles (37 km)? What am I missing here?

How to (partially) cross-divide 3D vectors? [migrated]

2024-08-09T18:33:20Z

A useful technique for vector cross-product:

It is often stated that there is no way to 3D cross-divide vectors, which is true, with caveats.

If we know that $ \vec{A} \times \vec{B} = \vec{C} $, and we know the value of $\vec{A}$ and $\vec{C}$, there are an infinite number of possible $\vec{B}$ vectors that satisfy $ \vec{A} \times \vec{B} = \vec{C} $.

The vector $\vec{B}$ can be always decomposed into a vector $\vec{B}_{parr}$ parallel to $\vec{A}$ ,

and a vector $\vec{B}_{p}$ perpendicular to $\vec{A}$.

$\vec{B}_{parr}$ contributes nothing to the value of $\vec{A} \times \vec{B}$.

$$ |\vec{B}_{p}| = |\vec{B}|sin(\theta) $$

$$ \vec{A} \times \vec{B} = \vec{A} \times \vec{B}_{p} $$

All vectors that can be expressed as $ \vec{B}_{p} + ( scalar * \vec{A} ) $ satisfy $ \vec{A} \times \vec{B} = \vec{C}$.

$$ \vec{A} \times ( \vec{B}_{p} + (scalar * \vec{A} ) ) $$

$$ = \vec{A} \times \vec{B}_{p} + ( scalar * \vec{A} \times \vec{A} ) $$

$$ = \vec{A} \times \vec{B}_{p} $$

Given $\vec{A}$ and $\vec{C}$, the full exact solution set for $\vec{B}$ is $\vec{B}_{p} + ( scalar * \vec{A} )$ .

Given only $\vec{A}$ and $\vec{C}$, we cannot determine which $\vec{B}$ (from the solution set) generated $\vec{C}$, but we can determine the unique $\vec{B}_{p}$.

We can define "vector (partial) cross-divide"

$$ \vec{C}//\vec{A} \equiv (\vec{C} \times \vec{A})/(\vec{A} \cdot \vec{A}) = (\vec{C} \times \hat{A})/|\vec{A}| = \vec{B}_p$$

You can see this method produce useful results here:

https://www.youtube.com/watch?v=_DkXD-cZLQE

derivation: $$ |\vec{C}//\vec{A}|= |(\vec{C} \times \vec{A})/(\vec{A} * \vec{A})| $$

$$ |\vec{C}| = |\vec{A}||\vec{B}|sin(\theta) = |\vec{A}||\vec{B}_{p}| $$

$$ |\vec{C}//\vec{A}|= |(|\vec{B}_{p}||\vec{A}||\vec{A})/(|\vec{A}||\vec{A}|)| $$

$$ |\vec{C}//\vec{A}|= |(\vec{B}_{p})| $$

$\vec{C}//\vec{A}$ and $\vec{B}_{p}$ have identical length, are both perpendicular to both $\vec{C}$ and $\vec{A}$, and the right hand rule points them in the same direction.

$$ \vec{B}_{p} = \vec{C}//\vec{A} $$

Conversely, $ \vec{A}_{perpendicular} = - \vec{C}//\vec{B} $.

Proof by component appears as an answer below.

If we want to apply the formula for torque $= pE\sin\theta$. It is given in the problem that theta~0° [closed]

2024-08-02T09:07:02Z

Why do we take sin theta to be theta only and not as sin0° ie., 0?

Position of a particle in the plane [closed]

2024-07-29T15:57:47Z

I'm here to ask a really stupid question just to be sure of its answer. My professor gave us an exercise where we have to determine the Lagrangian of a system that is formed by a circular ring of mass $M$ and radius $R$ which is placed in the $(x,z)$ plane. In this plane, it rotates around one of his point $O$ fixed in the origin of the axes. Also, there is a particle $P$ of mass $m$ that moves on this ring-like object without friction.

The Lagrangian has to be written in terms of the generalized coordinates $\theta$ and $\phi$, where $\theta$ is the angle formed by the segment that connects $O$ with the center $C$ of the ring guide with the $z$ axes and $\phi$ is the angle formed by the segment that connects the position of the particle $P$ with the center of the ring guide $C$ and the $z$ axes.

I have no problem in understanding how to write the Lagrangian itself, but I'm more confused on how the professor wrote the solution for this problem. They stated that the coordinates of the center $C$ of the guide can be written as: $$ \begin{cases} x= R\sin \theta\\ z=-R \cos\theta \end{cases} $$ where I would have written since $z = R \sin\left(\frac{\pi}{2}-\theta\right)=R \cos\theta$.

EDIT: I'll add an image of the problem so that it is easier to visualize the system. This is not official, it is how I understood it. Keep in mind that the ring guide can rotate around the origin $O$, so $\theta$ is not fixed.

Conception of earth's size based on the width of the observable horizon when standing at sea level and the circumference of the earth

2024-07-22T14:33:17Z

As an aid to conceiving of the size of the earth, using the information that the horizontal (left to right or right to left) width of (not the distance to) the observable horizon when standing at sea level is approximately 3.1 miles (5 kilometers) and the circumference is 24,901 miles (40,075 kilometers), can it be said that there are roughly slightly over 8,000 such horizons that fit into the circumference of the earth? I’m not sure if my colloquial, linear or arc notion of the observable horizon standing at a relatively “linear” beach is as correct a definition as one in which one pans one’s head 360 degrees. Actually, based on the distance to a lighthouse about 3 miles away on one side, the 3.1 mile distance for the observable width of the horizon seems a bit low -- by about half. (I got that 3.1 mile number off the internet). Perhaps there may be some perspective diminishment at play somehow.

I note the answer to a previously deleted, related question was:

If we are on a beach, and our eyes at 3m above the sea level, the horizon is a circle of about 6km. It seems consistent with our intuition.

Discovery of a Formula for Geostationary Orbit Distance: Seeking Expert Feedback [closed]

2024-07-17T19:45:26Z

I’m an amateur enthusiast without a formal academic background in mathematics or science. Recently, I stumbled upon an idea and derived a formula that I believe calculates the distance traveled along the geostationary orbit (GEO) as a function of the pointing angle (($\gamma$)) from a point on Earth’s surface. The formula seems to account for the fact that the observation point is not at the center of the Earth but on its surface.

Here’s the formula I came up with in Latex: $$ d = 2R \arcsin \left( \frac{\sqrt{(R-r)^2 + \left[ r \sin \left( \gamma + \frac{3\pi}{2} \right) + R \right]^2 - 2(R-r) \left[ r \sin \left( \gamma + \frac{3\pi}{2} \right) + R \right] \cos(\gamma)}}{2R} \right) $$ Where:

•    R  is the distance of the GEO from the center of the Earth,
•    r  is the radius of the Earth,
•    gamma is the pointing angle relative to the vertical direction.

I’ve tested the formula for various values of $\gamma$, and it seems to work well for the edge cases ($\gamma = 0 \text{ and } \gamma = \pi$), and it also shows an exponential-like behavior for angles close to $\pi/2$, which makes sense geometrically as the distance to the orbit increases.

I’m curious to know what more experienced mathematicians and physicists think about this. Is this formula known in the literature? Does it correctly account for the geometrical complexities involved?

Any feedback, corrections, or references to similar known formulas would be greatly appreciated. Thanks in advance for your insights!

Throwing a ball in the air

2024-07-10T14:24:31Z

When we throw a ball in the air, we know that if we do not throw it too high, then g can be held constant over the trajectory and we can approximate the curve by a parabola. However we also know that if we project a body at a height from the surface of the earth and the velocity is lower than the first cosmic velocity $v<(\frac {GM}{R})^{1/2}$ then the path followed by the body will be an ellipse. So then we can say that the trajectory is an ellipse. So in this case we approximated an ellipse by a parabola. Can we always do this?

Einstein's notion of "covariant"

2024-07-10T06:21:56Z

In his The Meaning of Relativity, pg. $11-12$, Einstein explains the notion of "covariant" along the following lines:
Consider a point $\mathbf x$ on a straight line $\mathbf x -\mathbf A=\lambda\mathbf B$ in $\mathbb R^3$. Without a loss of generality we can assume that $\mathbf B^T\mathbf B =1$. Now, consider an orthogonal linear transformation (OLT) from the $K_{\mathbf x}$ frame to $K'_{\mathbf x}$ defined by: $$\mathbf x'=\mathbf a +b\mathbf x$$ where $b=[b_{\mu\nu}]$ is the transformation matrix satisfying $b^Tb=I$. Then the straight line in our transformed frame becomes $\mathbf{x'-A'=}\lambda\mathbf B'$ with $\mathbf A'=b^T\mathbf A$ and $\mathbf B'=b^T\mathbf B$. It therefore implies that straight lines have an underlying property which is independent of the system of coordinates. Formally, this depends upon the fact that the (vector) quantity $\vec q :=(\mathbf x -\mathbf A)-\lambda\mathbf B$ is transformed as components of an interval, $\Delta\mathbf x$. If $\vec q=\vec 0$ for one system of Cartesian coordinates, then $\vec {q'}=\vec 0$ for all systems. Thus we can say that the equation of a straight line is "covariant" with respect to OLT's.

What I think of the assumption $\mathbf B^T\mathbf B =1$ is that here Einstein is referring to set of homothetic transformations centered at the point $\mathbf A,$ that an OLT maps a homothety centered at $\mathbf A$ to one centered at $\mathbf A'$, preserving their scaling factor $\lambda.$ So, straight lines are mapped to straight lines by an OLT, the points on which are therefore called "covariant" vectors. Is my interpretation correct?

Edit

After further reading I find that he is talking about the covariance of laws of physics expressed as equations: a law $F=0$ where $F$ may be a vector or scalar or tensor expression in general, if this law preserves its nature under a transformation, that is if it is transformed to $F'=0$ under the said transformation where $F'$ has similar mathematical form as $F$, then the law is said to be covariant with respect to that transformation.

Potential of circle and $n$-gon

2024-07-09T04:54:58Z

Consider power central fields $f \sim r^k, - \infty < k < \infty$, and unit circle, "charged" in the sense of the field.
We will be interested in the potential within the circle, i.e. scalar sum or integral.
The potential $\psi$ of a charge or small element will also be a power,
except $f \sim r^{-1}$, then $\psi \sim \ln (r)$

Var 1. The charge is distributed continuously.
It is known that for $f \sim r^{-2}, \psi \sim r^{-1}$ the force acting on the test charge inside the spheres are $0$ and the potential is constant. This is proved not only by Gauss’s theorem,
but also through the consideration of sites cut out by opposite bodily angles.
In the case of circle and other powers I find it difficult to apply the integral theorems.

Applying the idea of angles and arcs, it turns out that for $f \sim r^{-1}, \psi \sim \ln (r)$ the full potential inside the circle must be constant, and this is confirmed by the trial by calculations.
Question 1.1 How to prove it?
Question 1.2 Are there other central fields, perhaps not power law,
for which the potential inside the circle is constant?

Var 2. If we replace the circle with a right $n$-gon, the angle dependence appears.
But not always! As revealed intuitively, and now proven:

If $k = 2 d > 0$, the potential does not depend on the angle at $d < n$

Question 2.1 Can the linked reasoning be generalised and supplemented (eg for $k<0$)?
Question 2.2 Even in cases of other $k>0$, the potential dependence on the angle is very weak.
For example, at $k = 1, n = 3$ and moving the test charge around a radius of $0.5$
potential varies within $3.14 < \psi < 3.23$
And for $n = 4$ $4.23 < \psi < 4.25$
No "approximation" of a circle by a triangle and a square can be said. So why this is so?

Why does moment of inertia stop at 1/2 as solidness of a cylinder increases?

2024-07-09T01:14:58Z

So I have two things about moment of inertia:

The I for a hoop is $I=MR^2$ and the I for a solid disk is $I=\frac{1}{2}MR^2$

I've noticed that as "solidness" goes up from 0% (completely unsolid, hoop) to 100% (solid disk) the moment of inertia lowers until it's exactly 1/2 of the hoop (when it's solid).

Using the equation for a cylinder bound by $R_1$ (inner radius) and $R_2$ (outer radius) of $$I=\frac{1}{2}M(R_1^2+R_2^2)$$ we can think of "solidness" as the inverse percentage $R_1$ is of $R_2$.

I plotted this on a graph taking $R_1$ as the x-axis and moment of inertia I as the y-axis (and assumed $R_2=1$ and $M=1$):

The question I have is: why does moment of inertia stop at 1/2 as the disk becomes 100% solid? Is this something inherent to the geometry of circles where $I=\frac{1}{2}$ if there's a mass of 1 kg distributed equally over a circle of radius 1m? Why is it specifically $I=\frac{1}{2}$?

Also, why is the graph parabolic from 1/2? I know it's parabolic because it's based on the $I=\frac{1}{2}M(R_1^2+R_2^2)$ parabolic equation but intuitively why does I go up parabolically?

Finding Exterior Confining Pressure from Interior Pressure Point for a Solid Disk

2024-06-13T00:16:42Z

Essentially, I've been wrapping the pictured object tightly with string to exert a confinement pressure on its exterior. It's been difficult however to make a good estimate of how much pressure is exerted on the object. Theoretical models I've found of string wrapped around a cylinder/disk seem to be too simple for making precise measurements. Right now I'm using a pressure sensor to take measurements from the center of the object where the two interior pillars meet (the sensor doesn't work well with large surface areas). I can make precise measurements of the pressure within the pillars using the sensor, the pressure between the pillars clearly doesn't give the value of the pressure exerted on the exterior, as it doesn't match other setups I've tried or the models.

I've tried to work out how the pressure between the interior pillars relates to the confinement pressure. However, I don't have much confidence in my results.

Just in case this must be solved numerically, the dimensions are as follows:

Radius of the disk: 17.5 mm Radius of the hollow region: 11 mm Width/Height of disk: 14 mm Radius of central pillar contact surface: 4.5 mm Depth of lipped edge: 4.5 mm

Is obeying the parallelogram law of vector addition sufficient to make a physical quantity qualify as a vector?

2024-06-08T17:08:20Z

I know that obeying the parallelogram rule of vector addition is a necessary condition for vectors. But is it sufficient? In other words, can there be a quantity that is added using the method but does not have either of magnitude or direction?

The source of my confusion is that many physics texts list three properties that a physical quantity must have in order to be classified as a vector. It should: have a magnitude, have a direction and be added using the parallelogram rule of vector addition.

But I think that the third condition, in and of itself, is sufficient, i.e. the first two fall under its umbrella. So my question basically is -- are the magnitude and direction conditions in the (elementary, physics) definition of vectors redundant, if the obedience of the parallelogram rule has already been taken care of?

In my opinion, the answer is obviously yes, but one can never be too sure.

Why are spherical shapes so common in the universe?

2024-06-05T14:18:46Z

I have a simple question. Why are most objects in the observable universe spherical in shape? Why not conical, cubical, cuboidal for instance? I am furnishing a few points to justify this statement:

Most of the planets in the universe are spheroids.
Other heavenly bodies (e.g.: stars and moons) are also spherical.
Even elementary particles like atoms are best described as spheres (I admit that they don't have well-defined boundaries, but we still use terms like atomic radius when describing atoms.)

So, my primary question boils down to this single line:

Why is the spherical shape so ubiquitous in the observable universe (both in the microscopic and the macroscopic world)?

Describing force accumulation trend of an infinite volume with evenly distributed radiative sources

2024-05-30T01:38:38Z

I am looking for confirmation if I've built my equation properly.

My goal is to describe the change in force over time at a given point if evenly distributed radiators (in-phase or cumulative energy/force) in an effectively infinite volume begin emitting simultaneously. One might visualize this as a matrix of light bulbs, but I am strictly calculating for non-interfering radiation (no phase considerations).

Sir Isaac Newton’s inverse square law ($F=\frac{1}{r^2}$) describes the reduction of power from a given source over distance. But I am interested in the influence of multiple sources on a point in the center, so I'm looking at the geometry of progressive spheres with equal surface density of sources ($A=4r^2$).

Attempting to integrate the trend and apply it to gravitational force (the radiative force of interest, 'though any non-interfering radiation should work), my result is the following equation. NOTE: I have not fully simplified it intentionally to show the components used:

$F_{total}=\int_{r_{min}}^{r_{max}}G\frac{M(M4\pi r^2)}{r^2}dr$

I expect the change in the difference in the range of $r$ has a direct linear relationship to $F_{total}$. This should be the case for any set of $r_{min}$ and $r_{max}$ values. The trend I hope to model is, if all radiative sources simultaneously begin emitting, $r_{min}=0$ and $r_{max}=tc$ for a given period of time.

Again, I'm looking for constructive criticism of my approach and any advice on how to correct or further enhance this equation. Thank you.

Modified Special Geometry of SUSY Moduli Space

2024-05-25T10:19:40Z

It is known that the Coulomb branches of 5d $\mathcal{N}=1$ and 4d $\mathcal{N}=2$ SUSY (both have eight supercharges) satisfy special geometry. This means that there exists a holomorphic prepotential $\mathcal{F}$ such that the Coulomb branch is given by the hypersurface $$\mathcal{CB}=\{T^I,I=0,...,n_V\vert\mathcal{F}[T]=1\},$$ where $n_V$ denotes the number of vector multiplets.

It is also known that higher-derivative corrections to supergravities change this special geometry constraint, see e.g. this post.

$\textbf{Question:}$ How can this new, modified constraint be interpreted? Is the Coulomb branch still given by the hypersurface $\mathcal{F}[T]=1$ or by the new constraint $\mathcal{F}[T]=1+\mathcal{G}[T]$? If the latter is true, does this mean that higher-derivative corrections break special geometry and also the direct product structure of the SUSY moduli space?

$\textbf{Remark:}$ From geometric engineering (M-Theory on elliptically fibered CY$_3$), I would expect the former to be true. However, I am very unsure about this!

$\textbf{Edit:}$ Maybe I should be more explicit on why this correction concerns me. Consider M-Theory on a Calabi-Yau threefold $Y$. Then the Kähler moduli space is identified with the $\mathcal{N}=1$ Coulomb branch in five dimensions. The prepotential $\mathcal{F}$ is in this case simply the total volume $\mathcal{V}_Y$ of $Y$ which sits in a 5d hypermultiplet. Keeping $\mathcal{F}\equiv\mathcal{V}_Y=1$ fixed gives a constraint on the scalars of the vector multiplets (corresponding to the Kähler parameters of $Y$) and this constrained hypersurface is then the Coulomb branch. So far, so good. What I find worrying is that with the inclusion of $$S_{ARR}\sim\int_{\mathcal{M}_5}C_IA^I\wedge\mathrm{tr}R^2,$$ the condition $\mathcal{F}[T]=1$ is replaced by $\mathcal{F}[T]=1+\mathcal{G}[T]$ and it seems to me that now there is no clear distinction between Coulomb and Higgs branch possible anymore. This seems particularly strange as I always considered the direct product structure to be a very rigid condition coming from SUSY and seeing it broken due to a higher-derivative correction feels weird.

Is displacement vector always the shortest path?

2024-05-19T07:56:41Z

I read that the displacement vector of a particle is the shortest path between its initial and final positions since it's a straight line joining the two points, this holds true for me till a 2D plane but when we enter into 3D figures like a cube then it doesn't seem to always hold true there. So, what's the correct explanation?

Why are material properties often described by symmetric tensors?

2024-05-06T11:56:11Z

It seems to me that whenever there is some material parameter for a continuum, it is described by a symmetric tensor. This is the case for the mechanical stress tensor, permittivity of crystals, the inertia tensor, even higher-order nonlinear susceptibility represented by third- and higher order tensors.

On the other hand, I don't know any material properties described by general, nonsymmetric tensors.

Is there an underlying reason for this that is shared by the examples provided?

Does the top of a wheel really move at twice the velocity of the center?

2024-05-05T15:32:00Z

According to the physics of a wheel rolling without slipping, the topmost point moves twice as fast as the wheel.

But I tried an experiment:

Take a wheel on a table and hold a ruler horizontally in contact with the top of the wheel. Roll the wheel without slipping by pushing the ruler forward, such that the center of the wheel moves forward by $10 \,\text{cm}$.

But the ruler was supposed to move $20 \,\text{cm}$, isn't it? But it didn't, it just seemed to move $10 \,\text{cm}$. Am I interpreting something wrong?

Doubt in pitch of screw gauge [closed]

2024-05-04T12:39:37Z

I have just started learning about screw gauge and I came across this statement about pitch

The pitch is the distance between two consecutive threads of a screw which is equal to the distance moved by the screw due to one complete rotation of the cap.

I don't understand why it is so. Please explain in simple language.

Geometry of spacetime at different length scales [closed]

2024-04-26T15:21:55Z

Does spherical geometry govern physics at the quantum scale?

My motivation for this question came from studying non-Euclidean geometry. When we go down from general relativistic length scales to everyday length scales, geometry changes from non-Euclidean to Euclidean. Does this process continue as we move further down to quantum length scales? Does the shape of space continue to change? From a saddle to a flat sheet to a sphere, so to speak.

An example would be the 3 parallel lines that meet at the edge of the universe forming a triangle with sum of angles zero. As we shrink this triangle down the angle sum increases until at our scale it is a classical triangle. As this triangle shrinks down to a point does it's angle sum increase? Does it appear convex from our viewpoint and might this explain apparent faster than light travel from our viewpoint?

Is there such a thing as a "physical" fractal?

2024-04-26T06:50:57Z

The recent discovery of a molecule that mimics the Sierpinski gasket has spurred headlines identifying it as the first fractal scientists have found in nature. I find these claims highly dubious because it's either entirely impossible for a real fractal to be realized in nature or tons of other structures are trivially fractals even though we would normally not think about them as such.

The point of contention here is that, strictly speaking, a fractal must exhibit self-similarity over an infinite number of generations. No such structures are available outside of mathematical idealizations. We'd be tempted to say, then, that this molecule is not a fractal because it lacks the above property.

That being said, physicists are comfortable with calling space-time an infinitely differentiable manifold even though infinitely differentiable things are impossibilities. In this sense, we might shove infinities under the rug as physicists and say that the molecule is a fractal because it looks like the Sierpinski gasket over a finite number of generations. This would make it a prefractal, however. Moreover, if we're this lax about self-similarity, it follows that clouds, coastlines, and lightning must be fractals as well, in which case the Sierpinski molecule is obviously not the first fractal in nature

So, are there such things as physical fractals or not?

How do parallel reflected rays meet to form image at infinity? If they never meet then how is image formed?

2024-04-23T12:19:13Z

In my textbooks it is written that when an object is kept at focus, its image is formed at infinity and is real. But how is this possible because parallel lines never meet and it is necessary for rays to meet to form image.

Furthermore on youtube there are videos where the object is kept at focus of concave/convex mirror/lens respectively its image is formed very far. How is this posible as parallel rays never meet??

Please give me an answer in the context of geometric optics.

Average Speed in one half of an elliptical orbit

2024-04-13T15:37:22Z

I was wondering whether the average speed along one half of an elliptical orbit (say in a star planet system) had a closed form exact solution using Kepler's laws.

My approach was using the approximation for circumference of the ellipse, dividing by 2 and then dividing the entire thing by half the time period of an elliptical orbit. While this approach is straightforward, it only has an approximate answer.

How can I see Orion's Belt in winter and summer?

2024-04-13T10:07:14Z

How can the 23 degree tilt of the Earth enable someone in Argentina to see the same constellation (Orion's Belt) in winter as someone in Britain in summer?

How do we measure the position of a body? [duplicate]

2024-04-11T14:20:40Z

I am a high school student and I was studying kinematics about position of a body. So, one thing I do not understand in this diagram is the position of woman. Initally, at $t=0$, this woman was at $x=1.5\,\text{m}$ from reference frame; but her hands, her legs or any point of the woman are at different positions. So how we can say that this woman is at $x=1.5\,\text{m}$? How we can define the position of the woman?

Collection efficiency of mirror

2024-03-30T18:12:47Z

I want to compare plane and parabolic mirror for collection of photon from a point source. Intuitively, parabolic mirror focuses all rays from point source and gives parallel rays therefore has a higher collection efficiency. How do I mathematically prove it? Can someone help in estimating solid angle for both the cases? The mirrors have a height of 1.7mm and width of 4.3mm. The sample is placed around 1.05mm from mirror surface. And focal length in case of parabolic mirror is 0.75mm. How to calculate the efficiency analytically.

Does the term $d ( \omega_{ab} \wedge \theta^a \wedge \theta^b )$ have any significance?

2024-03-26T03:19:35Z

If $\omega_{ab}$ is the spin connection 1-form, and $\theta^a$ are the tetrad 1-forms, then one has the equality

\begin{equation} \int \, d ( \epsilon_{abcd} \omega^{ab} \wedge \theta^c \wedge \theta^d ) = \int d^3 x \sqrt{h} K \end{equation}

where $K$ is the extrinsic curvature scalar, and the second term is the GHY boundary term.

Is there a similar geometrical interpretation for the term

\begin{equation} \int d ( \omega_{ab} \wedge \theta^a \wedge \theta^b)? \end{equation}

Does this term have a name, or appear in any known theories? Is it zero?

One might also see similarities with these equations involving the Riemann curvature 2-form $R_{ab}$ \begin{equation} \epsilon_{abcd} R^{ab} \wedge \theta^c \wedge \theta^d = R \end{equation}

\begin{equation} R_{ab} \wedge \theta^a \wedge \theta^b = R_{\mu \nu \rho \sigma} \epsilon^{\mu \nu \rho \sigma} = 0. \end{equation}

The first term is the usual Einstein-Hilbert lagrangian whereas the second is the Holst term. Therefore, the term I am asking about is the analog of the Holst term for the GHY boundary term.

Edit: @Adversing has claimed in the comments that the term is zero in Einstein gravity. Does anyone have a proof of this fact?

Why is the refractive index for light rays travelling in circular paths proportional to $1/r$?

2024-03-22T19:11:51Z

While studying optics, I came across a problem with solution in which the trajectory of light rays was known—circular paths around a fixed point in space, and the question was that of determining the refraction index as a function of the distance $r$ from that fixed point (given that the refractive index at $r = r_0$ is $n_0$. A bit of a baffling question to me, as without at least knowing where the boundaries are Snell's law seemed to be utterly useless. The only thing that made sense to me was that the refractive index would depend solely on $r$, and the light rays were staying in the path with the least change in refractive index. Anyway the solution reads
$$n \cdot 2 \pi r = constant$$
$$n = {constant\over r}$$
$$n = {n_0r_0\over r}$$
At first, my intuition justified it as being similar to angular momentum (I understood it as light behaving as a particle), in that $v$ would also be proportional to $1/r$. But then eventually I realized $n$ is $c/v$, so in fact here the speed of light is proportional to $r$! The only thread I have at this point is Fermat's principle, but when the trajectory is known and not the refraction index I might as well be reconstructing a chess game by only seeing the final position. Can anyone make sense of this? (I don't suppose there is a nice "formula" for the refraction index function, as in like the one for the Lagrangian in mechanics; ${1\over2} mv^2 - V(x)$, is there?)